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Java/CGAL verify if a graph is connected (some constraints in description)

it's my first time with CGAL, some of you may argue why do I have to learn CGAL from something like that, but it's a new project that I must do (and... yes, I must use CGAL and Java combined) :/ Long story short... I only have:
Two double arrays, representing x and y coordinates of my vertices. Let's call them double[] x, y;.
Both arrays have S random values.
Two vertices, u and w are connected if distance(x[u], y[u], x[w], y[w]) < CONSTANT (ofc. I do distanceSquared(x[u], y[u], x[w], y[w]) < CONSTANT_SQUARED, so I avoid to call sqrt()).
x and y are filled randomly with values from 0 to UPPER_LIMIT, no other infos are given.
Question, do x and y describes a connected graph?
Right now I have two algoritms:
Algorithm 1:
Build adjacency list (Arraylist<Integer>[] adjLists;) for each vertex (only upper triangular matrix explored). Complexity O(|V|^2) (V = vertices set).
Recursive graph exploration, vertex marking and counting, if visited vertex equals S my graph have only one connected component, my graph is connected. Complexity O(|E|) (E = edges set).
Algorithm 2:
private static boolean algorithmGraph(double[] x, double[] y) {
int unchecked, inside = 0, current = 0;
double switchVar;
while (current <= inside && inside != S - 1) {
unchecked = inside + 1;
while (unchecked < S) {
if ((x[current] - x[unchecked]) * (x[current] - x[unchecked]) + (y[current] - y[unchecked]) * (y[current] - y[unchecked]) <= CONSTANT_SQUARED) {
inside++;
// switch x coordinates | unchecked <-> inside
switchVar = x[unchecked];
x[unchecked] = x[inside];
x[inside] = switchVar;
// switch y coordinates | unchecked <-> inside
switchVar = y[unchecked];
y[unchecked] = y[inside];
y[inside] = switchVar;
}
unchecked++;
}
current++;
}
return inside == S - 1;
}
Funny thing the second one is slower, I do not use data structures, the code is iterative and in-place but the heavy use of switch makes it slow as hell.
The problem spec changed and now I must do it with CGAL and Java, I'll read the whole "https://github.com/CGAL/cgal-swig-bindings" to learn how to use CGAL within Java.... but I'd like some help about this specific instance of CGAL code... Are there faster algorithms already implemented in CGAL?
Thank you for your times guys! Happy coding!

I believe that, without a method of spatial indexing, the best performance you are going to achieve in the worst-case-scenario (all connected) is going to be O(n*(n-1)/2).
If you can afford to build a spatial index (have enough memory to pay for the boost in speed), you may consider R-tree and variants - insertion is O(n) searching is O(log2(n)): this will get your "outlier detection by examining distances" approach for a cost of of O(n*log2(n)) in the worst-case-scenario.
A notable result

Algorithm to detect and combine overlapping / colliding circles

I'm trying to write a time efficient algorithm that can detect a group of overlapping circles and make a single circle in the "middle" of the group that will represent that group. The practical application of this is representing GPS locations over a map, put the conversion in to Cartesian co-ordinates is already handled so that's not relevant, the desired effect is that at different zoom levels clusters of close together points just appear as a single circle (that will have the number of points printed in the centre in the final version)
In this example the circles just have a radius of 15 so the distance calculation (Pythagoras) is not being square rooted and compared to 225 for the collision detection. I was trying anything to shave off time, but the problem is this really needs to happen very quickly becasue it's a user facing bit of code that needs to be snappy and good looking.
I've given this a go and I it works with small data sets pretty well. 2 big problems, it takes too long and it can run out of memory if all the points are on top of one another.
The route I've taken is to calculate distance between each point in a first pass, and then take the shortest distance first and start to combine from there, anything that's been combined becomes ineligible for combination on that pass, and the whole list is passed back around to the distance calculations again until nothing changes.
To be honest I think it needs a radical shift in approach and I think it's a little beyond me. I've re factored my code in to one class for ease of posting and generated random points to give an example.
package mergepoints;
import java.awt.Point;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Merger {
public static void main(String[] args) {
Merger m = new Merger();
m.subProcess(m.createRandomList());
}
private List<Plottable> createRandomList() {
List<Plottable> points = new ArrayList<>();
for (int i = 0; i < 50000; i++) {
Plottable p = new Plottable();
p.location = new Point((int) Math.floor(Math.random() * 1000),
(int) Math.floor(Math.random() * 1000));
points.add(p);
}
return points;
}
private List<Plottable> subProcess(List<Plottable> visible) {
List<PlottableTuple> tuples = new ArrayList<PlottableTuple>();
// create a tuple to store distance and matching objects together,
for (Plottable p : visible) {
PlottableTuple tuple = new PlottableTuple();
tuple.a = p;
tuples.add(tuple);
}
// work out each Plottable relative distance from
// one another and order them by shortest first.
// We may need to do this multiple times for one set so going in own
// method.
// this is the bit that takes ages
setDistances(tuples);
// Sort so that smallest distances are at the top.
// parse the set and combine any pair less than the smallest distance in
// to a combined pin.
// any plottable thats been combine is no longer eligable for combining
// so ignore on this parse.
List<PlottableTuple> sorted = new ArrayList<>(tuples);
Collections.sort(sorted);
Set<Plottable> done = new HashSet<>();
Set<Plottable> mergedSet = new HashSet<>();
for (PlottableTuple pt : sorted) {
if (!done.contains(pt.a) && pt.distance <= 225) {
Plottable merged = combine(pt, done);
done.add(pt.a);
for (PlottableTuple tup : pt.others) {
done.add(tup.a);
}
mergedSet.add(merged);
}
}
// if we haven't processed anything we are done just return visible
// list.
if (done.size() == 0) {
return visible;
} else {
// change the list to represent the new combined plottables and
// repeat the process.
visible.removeAll(done);
visible.addAll(mergedSet);
return subProcess(visible);
}
}
private Plottable combine(PlottableTuple pt, Set<Plottable> done) {
List<Plottable> plottables = new ArrayList<>();
plottables.addAll(pt.a.containingPlottables);
for (PlottableTuple otherTuple : pt.others) {
if (!done.contains(otherTuple.a)) {
plottables.addAll(otherTuple.a.containingPlottables);
}
}
int x = 0;
int y = 0;
for (Plottable p : plottables) {
Point position = p.location;
x += position.x;
y += position.y;
}
x = x / plottables.size();
y = y / plottables.size();
Plottable merged = new Plottable();
merged.containingPlottables.addAll(plottables);
merged.location = new Point(x, y);
return merged;
}
private void setDistances(List<PlottableTuple> tuples) {
System.out.println("pins: " + tuples.size());
int loops = 0;
// Start from the first item and loop through, then repeat but starting
// with the next item.
for (int startIndex = 0; startIndex < tuples.size() - 1; startIndex++) {
// Get the data for the start Plottable
PlottableTuple startTuple = tuples.get(startIndex);
Point startLocation = startTuple.a.location;
for (int i = startIndex + 1; i < tuples.size(); i++) {
loops++;
PlottableTuple compareTuple = tuples.get(i);
double distance = distance(startLocation, compareTuple.a.location);
setDistance(startTuple, compareTuple, distance);
setDistance(compareTuple, startTuple, distance);
}
}
System.out.println("loops " + loops);
}
private void setDistance(PlottableTuple from, PlottableTuple to,
double distance) {
if (distance < from.distance || from.others == null) {
from.distance = distance;
from.others = new HashSet<>();
from.others.add(to);
} else if (distance == from.distance) {
from.others.add(to);
}
}
private double distance(Point a, Point b) {
if (a.equals(b)) {
return 0.0;
}
double result = (((double) a.x - (double) b.x) * ((double) a.x - (double) b.x))
+ (((double) a.y - (double) b.y) * ((double) a.y - (double) b.y));
return result;
}
class PlottableTuple implements Comparable<PlottableTuple> {
public Plottable a;
public Set<PlottableTuple> others;
public double distance;
#Override
public int compareTo(PlottableTuple other) {
return (new Double(distance)).compareTo(other.distance);
}
}
class Plottable {
public Point location;
private Set<Plottable> containingPlottables;
public Plottable(Set<Plottable> plots) {
this.containingPlottables = plots;
}
public Plottable() {
this.containingPlottables = new HashSet<>();
this.containingPlottables.add(this);
}
public Set<Plottable> getContainingPlottables() {
return containingPlottables;
}
}
}

Map all your circles on a 2D grid first. You then only need to compare the circles in a cell with the other circles in that cell and in it's 9 neighbors (you can reduce that to five by using a brick pattern instead of a regular grid).
If you only need to be really approximate, then you can just group all the circles that fall into a cell together. You will probably also want to merge cells that only have a small number of circles together with there neighbors, but this will be fast.

This problem is going to take a reasonable amount of computation no matter how you do it, the question then is: can you do all the computation up-front so that at run-time it's just doing a look-up? I would build a tree-like structure where each layer is all the points that need to be drawn for a given zoom level. It takes more computation up-front, but at run-time you are simply drawing a list of point, fast.
My idea is to decide what the resolution of each zoom level is (ie at zoom level 1 points closer than 15 get merged; at zoom level 2 points closer than 30 get merged), then go through your points making groups of points that are within the 15 of each other and pick a point to represent group that group at the higher zoom. Now you have a 2 layer tree. Then you pass over the second layer grouping all points that are within 30 of each other, and so on all the way up to your highest zoom level. Now save this tree structure to file, and at run-time you can very quickly change zoom levels by simply drawing all points at the appropriate tree level. If you need to add or remove points, that can be done dynamically by figuring out where to attach them to the tree.
There are two downsides to this method that come to mind: 1) it will take a long time to compute the tree, but you only have to do this once, and 2) you'll have to think really carefully about how you build the tree, based on how you want the groupings to be done at higher levels. For example, in the image below the top level may not be the right grouping that you want. Maybe instead building the tree based off the previous layer, you always want to go back to the original points. That said, some loss of precision always happens when you're trying to trade-off for faster run-time.
EDIT
So you have a problem which requires O(n^2) comparisons, you say it has to be done in real-time, can not be pre-computed, and has to be fast. Good luck with that.
Let's analyze the problem a bit; if you do no pre-computation then in order to decide which points can be merged you have to compare every pair of points, that's O(n^2) comparisons. I suggested building a tree before-hand, O(n^2 log n) once, but then runtime is just a lookup, O(1). You could also do something in between where you do some work before and some at run-time, but that's how these problems always go, you have to do a certain amount of computation, you can play games by doing some of it earlier, but at the end of the day you still have to do the computation.
For example, if you're willing to do some pre-computation, you could try keeping two copies of the list of points, one sorted by x-value and one sorted by y-value, then instead of comparing every pair of points, you can do 4 binary searches to find all the points within, say, a 30 unit box of the current point. More complicated so would be slower for a small number of points (say <100), but would reduce the overall complexity to O(n log n), making it faster for large amounts of data.
EDIT 2
If you're worried about multiple points at the same location, then why don't you do a first pass removing the redundant points, then you'll have a smaller "search list"
list searchList = new list()
for pt1 in points :
boolean clean = true
for pt2 in searchList :
if distance(pt1, pt2) < epsilon :
clean = false
break
if clean :
searchList.add(pt1)
// Now you have a smaller list to act on with only 1 point per cluster
// ... I guess this is actually the same as my first suggestion if you make one of these search lists per zoom level. huh.
EDIT 3: Graph Traversal
A totally new approach would be to build a graph out of the points and do some sort of longest-edge-first graph traversal on them. So pick a point, draw it, and traverse its longest edge, draw that point, etc. Repeat this until you come to a point which doesn't have any untraversed edges longer than your zoom resolution. The number of edges per point gives you an easy way to tradeoff speed for correctness. If the number of edges per point was small and constant, say 4, then with a bit of cleverness you could build the graph in O(n) time and also traverse it to draw points in O(n) time. Fast enough to do it on the fly with no pre-computation.

Just a wild guess and something that occurred to me while reading responses from others.
Do a multi-step comparison. Assume your combining distance at the current zoom level is 20 meters. First, subtract (X1 - X2). If This is bigger than 20 meters then you are done, the points are too far. Next, subtract (Y1 - Y2) and do the same thing to reject combining the points.
You could stop here and be happy if you are good with using only horizontal/vertical distances as your metric for combining. Much less math (no squaring or square roots). Pythagoras wouldn't be happy but your users might.
If you really insist on exact answers, do the two subtraction/comparison steps above. If the points are within horizontal and vertical limits, THEN you do the full Pythagoras check with square roots.
Assuming all your points are not highly clustered very close to the combining limit, this should save some CPU cycles.
This is still approximately an O(n^2) technique, but the math should be simpler. If you have the memory, you could store distances between each set of points and then you never have to compute it again. This could take up more memory than you have and also grows at a rate of approximately O(n^2), so be careful.
Also, you could make a linked list or sorted array of all your points, sorted in order of increasing X or increasing Y. (I don't think you need both, just one). Then walk through the list in sorted order. For each point, check the neighbors out until (X1 - X2) is bigger than your combining distance. and then stop. You don't have to compare each set of points for O(N^2), you only have to compare neighbors that are close in one dimension to quickly prune your large list to a small one. As you move through the list, you only have to compare points that have a bigger X than your current candidate, because you already compared and combined with all previous X values. This gets you closer to the O(n) complexity you want. Of course, you would need to check the Y dimension and fully qualify the points to be combined before you actually do it. Don't just use the X distance to make your combining decision.

How do I know that my neural network is being trained correctly

I've written an Adaline Neural Network. Everything that I have compiles, so I know that there isn't a problem with what I've written, but how do I know that I have to algorithm correct? When I try training the network, my computer just says the application is running and it just goes. After about 2 minutes I just stopped it.
Does training normally take this long (I have 10 parameters and 669 observations)?
Do I just need to let it run longer?
Hear is my train method
public void trainNetwork()
{
int good = 0;
//train until all patterns are good.
while(good < trainingData.size())
{
for(int i=0; i< trainingData.size(); i++)
{
this.setInputNodeValues(trainingData.get(i));
adalineNode.run();
if(nodeList.get(nodeList.size()-1).getValue(Constants.NODE_VALUE) != adalineNode.getValue(Constants.NODE_VALUE))
{
adalineNode.learn();
}
else
{
good++;
}
}
}
}
And here is my learn method
public void learn()
{
Double nodeValue = value.get(Constants.NODE_VALUE);
double nodeError = nodeValue * -2.0;
error.put(Constants.NODE_ERROR, nodeError);
BaseLink link;
int count = inLinks.size();
double delta;
for(int i = 0; i < count; i++)
{
link = inLinks.get(i);
Double learningRate = value.get(Constants.LEARNING_RATE);
Double value = inLinks.get(i).getInValue(Constants.NODE_VALUE);
delta = learningRate * value * nodeError;
inLinks.get(i).updateWeight(delta);
}
}
And here is my run method
public void run()
{
double total = 0;
//find out how many input links there are
int count = inLinks.size();
for(int i = 0; i< count-1; i++)
{
//grab a specific link in sequence
BaseLink specificInLink = inLinks.get(i);
Double weightedValue = specificInLink.weightedInValue(Constants.NODE_VALUE);
total += weightedValue;
}
this.setValue(Constants.NODE_VALUE, this.transferFunction(total));
}
These functions are part of a library that I'm writing. I have the entire thing on Github here. Now that everything is written, I just don't know how I should go about actually testing to make sure that I have the training method written correctly.
I asked a similar question a few months ago.

Ten parameters with 669 observations is not a large data set. So there is probably an issue with your algorithm. There are two things you can do that will make debugging your algorithm much easier:
Print the sum of squared errors at the end of each iteration. This will help you determine if the algorithm is converging (at all), stuck at a local minimum, or just very slowly converging.
Test your code on a simple data set. Pick something easy like a two-dimensional input that you know is linearly separable. Will your algorithm learn a simple AND function of two inputs? If so, will it lean an XOR function (2 inputs, 2 hidden nodes, 2 outputs)?

You should be adding debug/test mode messages to watch if the weights are getting saturated and more converged. It is likely that good < trainingData.size() is not happening.
Based on Double nodeValue = value.get(Constants.NODE_VALUE); I assume NODE_VALUE is of type Double ? If that's the case then this line nodeList.get(nodeList.size()-1).getValue(Constants.NODE_VALUE) != adalineNode.getValue(Constants.NODE_VALUE) may not really converge exactly as it is of type double with lot of other parameters involved in obtaining its value and your convergence relies on it. Typically while training a neural network you stop when the convergence is within an acceptable error limit (not a strict equality like you are trying to check).
Hope this helps

Steps to perform document clustering using k-means algorithm in java

I need steps to perform document clustering using k-means algorithm in java.
It will be very useful for me to provide the steps easily.
Thanks in advance.

You need to count the words in each document and make a feature generally called bag of words. Before that you need to remove stop words(very common but not giving much information like the, a etc). You can generally take top n common words from your document. Count the frequency of these words and store them in n dimensional vector.
For distance measure you can use cosine vector.
Here is a simple algorithm for 2 mean for 1 dimensional data points. you can extend it to k mean and n dimensional data point easily. Let me know if you want n dim implementation.
double[] x = {1,2,2.5,3,3.5,4,4.5,5,7,8,8.5,9,9.5,10};
double[] center = new int[2];
double[] precenter = new int[2];
ArrayList[] cluster = new ArrayList[2];
//generate 2 random number from 0 to x.length without replacement
int rand = new int[2];
Random rand = new Random();
rand[0] = rand.nextInt(x.length + 1);
rand[1] = rand.nextInt(x.length + 1);
while(rand[0] == rand[1] ){
rand[1] = rand.nextInt(x.length + 1);
}
center[0] = x[rand[0]];
center[1] = x[rand[1]];
//there is a better way to generate k random number (w/o replacement) just search.
do{
cluster[0].clear();
cluster[1].clear();
for(int i = 0; i < x.length; ++i){
if(abs(x[i]-center1[0]) <= abs(x[i]-center1[1])){
cluster[0].add(x[i]);
}
else{
cluster[0].add(x[i]);
}
precenter[0] = center[0];
precenter[1] = center[1];
center[0] = mean(cluster[0]);
center[1] = mean(cluster[1]);
}
} while(precenter[0] != center[0] && precenter[1] != center[1]);
double mean(ArrayList list){
double mean = 0;
double sum = 0;
for(int index=0;index
}
The cluster[0] and cluster [1] contain points in the clusters and center[0], center[1] are the 2 means.
you need to do some debugging because I have written the code in R and just converted it into java for you :)

Does this help you? Also the wiki article has some links to implementations in other languages ready to be ported to java.
Steps of the algorithm:
Define the number of clusters you want to have
Distribute the points radomly in your problem space.
Link every observation to the nearest point.
calculate the center of mass for each cluster and place the point into the middle.
Link the points again to the centerpoints and repeat until the points dont move any more.

What do you want to cluster the documents based on? If it's by similarity you'll need to do some natural language processing first, and then you'll need a metric (some kind of assignment algorithm) to place the documents into clusters (crp works and is relatively straight forward).
The hardest part will be the NLP (language processing) if you're not clustering them based on something like "length". I can provide more info on all of these, but I won't dive down the rabbit hole if you don't need it.

Graph Theory: Find the Jordan center?

I'm trying to find the set of vertices that minimizes their distance to other vertices on a weighted graph. Based on a cursory wikipedia search, I think that this is called the Jordan Center. What are some good algorithms for finding it?
Right now, my plan is to get a list of the weight for each branch emanating from a given vertex. The vertices whose weights have the smallest relative difference will be the central ones. Any other ideas?
I'm using Java, but helpful answers don't necessarily need to be Java specific.

I woluld first use Dijkstra algorithm (it has to be run for each verticle) for computng shortest distances between all pairs of verticles - there are also some more efficient algorithms for that like Floyd-Warshall. Then for each verticle V you have to find Vm - the largest distance to any other verticles amongs the data retuirned form Dijkstra algorithm. Then, the verticles with the smallest Vm are the one in the graph center. Pseudocode:
int n = number of verticles;
int[][] D = RunDijkstraOrWarshall()
// D[a,b] = length of shortest path from a to b
int[] Vm = new int[n];
for(int i=0; i<n i++)
{
Vm[i] = 0
for(int j=0; j<n; j++)
{
if (Vm[i] < D[i,j]) Vm[i] = D[i,j];
}
}
minVm = int.Max;
for(int i=0; i<n ;i++)
{
if (minVm < Vm[i]) minVm = Vm[i];
}
for(int i=0; i<n ;i++)
{
if (Vm[i] == minVm)
{
// graph center contans i
}
}

Three algorithms for graph center problem are presented in this MSc thesis: A distributed algorithm for the graph center problem.

Starting from JGraphT version 1.1.0, you can simply use the method GraphMeasurer.getGraphCenter(). The underlying code uses a shortest path method. The user can choose which method to use, depending on some characteristics of the graph (e.g. sparse/dense/...).